Given a collection of candidate numbers (C)
and a target number (T),
find all unique combinations in C where
the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5
and
target 8
,
A solution set is:
[1,
7]
[1,
2, 5]
[2,
6]
[1,
1, 6]
class Solution {
public:
void comb(vector<int> &num, vector<int> &count, vector< vector<int> > &result,
vector<int> &record, int target, int sum, int pos, int s)
{
if (sum == target)
{
result.push_back(vector<int>());
for (int i = 0; i < pos; ++i)
result.back().push_back(record[i]);
}
if (pos >= record.size() || sum > target)
return;
for(int i = s; i < num.size(); ++i)
{
if (count[i] > 0)
{
count[i]--;
record[pos] = num[i];
comb(num, count, result, record, target, sum+num[i], pos+1, i);
count[i]++;
}
}
}
vector<vector<int> > combinationSum2(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int nSize = num.size();
vector< vector<int> > result;
if (nSize == 0) return result;
sort(num.begin(), num.end());
vector<int> nums;
nums.push_back(num[0]);
vector<int> count;
count.push_back(1);
for(int i = 1; i < num.size(); ++i)
{
if (num[i] == nums.back())
++count.back();
else
{
nums.push_back(num[i]);
count.push_back(1);
}
}
vector<int> record(nSize);
comb(nums, count, result, record, target, 0, 0, 0);
return result;
}
};